Just by looking at the question you know that LiOH is in excess. Thus, using n = cv, n(excess LiOH) = n(LiOH total) - n(H2SO4) = 0.0035 mol.
Since LiOH --> Li+ + OH-, n(OH) = 0.0035 mol.
To find [OH] = 0.0035/0.08 = 0.04375 mol/L.
pOH = -log(0.04375) = 1.359.
pH = 14 - pOH = 12.64.
Therefore the answer is D.
Neutralisation works the same with polyprotic acids, it just changes the mole ratios im pretty sure. However, for the concentrations, if a weak acid was in excess im pretty sure they would give more information for equlibrium calculations.